Probability of choosing a biased coin $C$ which has probability $3/15$ of getting heads, assuming we got head...












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$begingroup$


Full question: there are 3 biased coins $A$, $B$, and $C$ each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability $1/4$ for $A$, $1/4$ for $B$, and $1/2$ for $C$ of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin $C$?



My approach: Since the coin picked was $C$ and the result was head we merely multiply the probability of both those things happening concerning $C$:
$$frac1{15} cdot frac12 = frac1{30}$$






probability discrete-mathematics conditional-probability







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    2












    $begingroup$


    Full question: there are 3 biased coins $A$, $B$, and $C$ each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability $1/4$ for $A$, $1/4$ for $B$, and $1/2$ for $C$ of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin $C$?



    My approach: Since the coin picked was $C$ and the result was head we merely multiply the probability of both those things happening concerning $C$:
    $$frac1{15} cdot frac12 = frac1{30}$$






    probability discrete-mathematics conditional-probability







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Full question: there are 3 biased coins $A$, $B$, and $C$ each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability $1/4$ for $A$, $1/4$ for $B$, and $1/2$ for $C$ of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin $C$?



      My approach: Since the coin picked was $C$ and the result was head we merely multiply the probability of both those things happening concerning $C$:
      $$frac1{15} cdot frac12 = frac1{30}$$






      probability discrete-mathematics conditional-probability







      share|cite|improve this question











      $endgroup$




      Full question: there are 3 biased coins $A$, $B$, and $C$ each with probability $5/15, 3/15, 1/15$ of getting heads respectively. Also, they have probability $1/4$ for $A$, $1/4$ for $B$, and $1/2$ for $C$ of getting picked. If a coin was picked and tossed and the result was heads, what is the probability that the coin was coin $C$?



      My approach: Since the coin picked was $C$ and the result was head we merely multiply the probability of both those things happening concerning $C$:
      $$frac1{15} cdot frac12 = frac1{30}$$






      probability discrete-mathematics conditional-probability




      probability discrete-mathematics conditional-probability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 24 '18 at 7:49









      Asaf Karagila♦

      303k32429760




      303k32429760










      asked Dec 24 '18 at 4:50









      hussain sagarhussain sagar

      806




      806






















          2 Answers
          2






          active

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          3












          $begingroup$

          An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:




          1. Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.

          2. Draw $B$: $1/4 times 3/15 = 1/20$.

          3. Draw $C$: $1/2 times 1/15 = 1/30$.


          Thus, the chance that $C$ was drawn is
          $$
          frac{1/30}{1/12 + 1/20 + 1/30}
          = frac{1}{5/2+3/2 + 1}
          = frac15.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, this makes a lot of sense.
            $endgroup$
            – hussain sagar
            Dec 24 '18 at 4:57










          • $begingroup$
            @hussainsagar you are welcome
            $endgroup$
            – gt6989b
            Dec 24 '18 at 4:58



















          1












          $begingroup$

          What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.



          Guide:
          Use Bayes rule, that is



          $$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$






          share|cite|improve this answer









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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

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            3












            $begingroup$

            An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:




            1. Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.

            2. Draw $B$: $1/4 times 3/15 = 1/20$.

            3. Draw $C$: $1/2 times 1/15 = 1/30$.


            Thus, the chance that $C$ was drawn is
            $$
            frac{1/30}{1/12 + 1/20 + 1/30}
            = frac{1}{5/2+3/2 + 1}
            = frac15.
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you, this makes a lot of sense.
              $endgroup$
              – hussain sagar
              Dec 24 '18 at 4:57










            • $begingroup$
              @hussainsagar you are welcome
              $endgroup$
              – gt6989b
              Dec 24 '18 at 4:58
















            3












            $begingroup$

            An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:




            1. Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.

            2. Draw $B$: $1/4 times 3/15 = 1/20$.

            3. Draw $C$: $1/2 times 1/15 = 1/30$.


            Thus, the chance that $C$ was drawn is
            $$
            frac{1/30}{1/12 + 1/20 + 1/30}
            = frac{1}{5/2+3/2 + 1}
            = frac15.
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you, this makes a lot of sense.
              $endgroup$
              – hussain sagar
              Dec 24 '18 at 4:57










            • $begingroup$
              @hussainsagar you are welcome
              $endgroup$
              – gt6989b
              Dec 24 '18 at 4:58














            3












            3








            3





            $begingroup$

            An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:




            1. Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.

            2. Draw $B$: $1/4 times 3/15 = 1/20$.

            3. Draw $C$: $1/2 times 1/15 = 1/30$.


            Thus, the chance that $C$ was drawn is
            $$
            frac{1/30}{1/12 + 1/20 + 1/30}
            = frac{1}{5/2+3/2 + 1}
            = frac15.
            $$






            share|cite|improve this answer









            $endgroup$



            An intuitive argument based on Bayes' Theorem, says that getting heads was possible in one of three different ways:




            1. Draw $A$ with probability $1/4$ and flip heads with probability $5/15$, total chance of $1/4 times 5/15 = 1/12$.

            2. Draw $B$: $1/4 times 3/15 = 1/20$.

            3. Draw $C$: $1/2 times 1/15 = 1/30$.


            Thus, the chance that $C$ was drawn is
            $$
            frac{1/30}{1/12 + 1/20 + 1/30}
            = frac{1}{5/2+3/2 + 1}
            = frac15.
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 24 '18 at 4:55









            gt6989bgt6989b

            33.9k22455




            33.9k22455












            • $begingroup$
              Thank you, this makes a lot of sense.
              $endgroup$
              – hussain sagar
              Dec 24 '18 at 4:57










            • $begingroup$
              @hussainsagar you are welcome
              $endgroup$
              – gt6989b
              Dec 24 '18 at 4:58


















            • $begingroup$
              Thank you, this makes a lot of sense.
              $endgroup$
              – hussain sagar
              Dec 24 '18 at 4:57










            • $begingroup$
              @hussainsagar you are welcome
              $endgroup$
              – gt6989b
              Dec 24 '18 at 4:58
















            $begingroup$
            Thank you, this makes a lot of sense.
            $endgroup$
            – hussain sagar
            Dec 24 '18 at 4:57




            $begingroup$
            Thank you, this makes a lot of sense.
            $endgroup$
            – hussain sagar
            Dec 24 '18 at 4:57












            $begingroup$
            @hussainsagar you are welcome
            $endgroup$
            – gt6989b
            Dec 24 '18 at 4:58




            $begingroup$
            @hussainsagar you are welcome
            $endgroup$
            – gt6989b
            Dec 24 '18 at 4:58











            1












            $begingroup$

            What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.



            Guide:
            Use Bayes rule, that is



            $$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.



              Guide:
              Use Bayes rule, that is



              $$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.



                Guide:
                Use Bayes rule, that is



                $$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$






                share|cite|improve this answer









                $endgroup$



                What you computed is the probability that coin $C$ is chosen and you get a head $H$, that is $P(H cap C)$. It is not equalty to $P(C|H)$.



                Guide:
                Use Bayes rule, that is



                $$P(C|H)= frac{P(H|C)P(C)}{P(H)}=frac{P(H|C)P(C)}{P(Hcap A)+P(H cap B)+P(H cap C)}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 24 '18 at 4:55









                Siong Thye GohSiong Thye Goh

                100k1466117




                100k1466117






























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